Transcript SQC_Module7_DD-done

Statistical Quality Control in Textiles
Module 7:
Acceptance Sampling Techniques
Dr. Dipayan Das
Assistant Professor
Dept. of Textile Technology
Indian Institute of Technology Delhi
Phone: +91-11-26591402
E-mail: [email protected]
Introduction
Why Acceptance Sampling? [1]
Input
Yes
Manufacturing
Process
No
Scrap or
Rework
No
Yes
Customer
Is output
acceptable?
Is process
under
control?
Correction
Is input
acceptable?
Yes
Output
Whether the input or output materials are
acceptable or not can be found through a
technique called Acceptance Sampling.
Acceptance Sampling: Attributes & Variables
The input or output articles are available in lots or batches
(population). It is practically impossible to check each and every
article of a batch. So we randomly select a few articles (sample) from
a batch, inspect them, and then draw conclusion whether the batch is
acceptable or not. This is called acceptance sampling.
Sometimes the articles inspected are merely classified as defective or
non-defective. Then we deal with acceptance sampling of attributes.
Sometimes the property of the articles inspected is actually measured,
then we deal with acceptance sampling of variables.
Acceptance Sampling of Attributes
Definition
Let us take a sample of size n randomly from a batch. If the number
of defective articles in the sample is not greater than a certain number
c, then the batch is accepted; otherwise, it is rejected.
This is how we define acceptance sampling plan.
Probability of Acceptance
Let us assume that the proportion of defective articles in the batch is
p. Then, when a single article is randomly chosen from a batch, the
probability that it will be defective is p. Further assume that the batch
size is sufficiently larger than the sample size n so that this
probability is the same for each article in the sample. Thus, the
probability of finding exactly r number of defective articles in a
sample of size n is
n!
nr
r
P r  
p 1  p 
r ! n  r !
Now the batch will be accepted if r ≤ c, i.e., if r=0 or 1 or 2 or….or c.
Then, according to the addition rule of probability, the probability of
accepting the batch is
Operating Characteristic
Pa  p   P  r  0   P  r  1  P  r  2  
This tells that once n and c are
known, the probability of
accepting a batch depends
only on the proportion of
defectives in the batch. Thus,
Pa(p) is a function of p. This
function
is
known
as
Operating
Characteristic
(OC) of the sampling plan.
r c
n!
nr
r
 P r  c  
p 1  p 
r  0 r ! n  r  !
Pa  p 
Ideal OC Curve
1
Practical OC Curve
p
Acceptable Quality Level (AQL) p=p1
This represents the poorest level
of quality for the producer’s
process that the consumer would
consider to be acceptable as
process average. Ideally, the
producer should try to produce
lots of quality better than p1.
Assume
there
is
a
high
probability, say 1-, of accepting
a batch of quality p1. Then, the
probability of rejecting a batch of
quality p1 is , which is known as
producer’s risk.
When p  p1
then Pa  p1   1  .
Pa  p 
1
1 

p1
p
Lot Tolerance Proportion Defectives (LTPD) p=p2>p1
This represents the poorest level
of quality that the consumer is
willing to accept in an individual
lot. Below this level, it is
unacceptable to the consumer. In
spite of this, let there will be a
small chance (probability)  of
accepting such a bad batch by the
consumer, 
is
consumer’s risk.
When
known
as
p  p2 then Pa  p2   .
LTPD is also known as rejectable
quality level (RQL).
Pa  p 
1
1 



p1
p2
p
Finding of n and c
r c
n!
nr
r
Pa  p1   1    
p1 1  p1 
r  0 r ! n  r  !
r c
n!
nr
p2r 1  p2 
r  0 r ! n  r  !
Pa  p2     
Solution is based on 2 distribution with 2(c+1) degree of freedom
 22 c 1,1  2np1 
 
2
 2 c 1,  2np2 
22 c 1,1
22 c 1,
2np1 p1



2np2 p2
In a sampling plan, p1, p2, , and  are given. Then the value of c
can be found out from Equation , and then the value of n can be
found out from Equation .
Illustration [2]
Design a sampling plan for which AQL is 2%, LTPD is 5%, and the
producer’s risk and consumer’s risk are both 5%.
Here p1  0.02, p2  0.05,   0.05,   0.05.
For 2(c+1)=26 degree of freedom
Hence, c=12, then
226,0.95
226,0.05

22 c 1,1
22 c 1,
226,0.95  15.38
15.38
 0.3955  0.4
38.89

&
2np1 p1 0.02


 0.4
2np2 p2 0.05
226,0.05  38.89
Illustration (Contd.)
 0.02
 0.05
 226,0.95  15.38  2n p1  0.04n
or, n 
15.38
 384.5
0.04
 226,0.05  38.89  2n p2  0.1n
or, n 
38.89
 388.9
0.1
1
n   384.5  388.9   386.7  387
2
Effect of Sample Size n on OC Curve
pa  p 
As the sample size n
increases, the OC curve
becomes
more
like
idealized OC curve.
n  100, c  1
n  200, c  2
Plans with higher value
of
n
offers
more
discriminatory power.
n  300, c  3
n  400, c  4
n  500, c  5
p
(Note
that
c
proportional to n).
is
Effect of Acceptance Number c on OC Curve
pa  p 
As the acceptance number c
decreases, the OC curve
shifts to the left, however,
the slope of the curve does
not change appreciably.
n  100, c  5
n  100, c  4
n  100, c  3
n  100, c  2
n  100, c  1
n  100, c  0
p
Plans with smaller value of c
provide discrimination at
lower levels of lot fraction
defective than do plans with
larger values of c.
Acceptance Sampling of Variables
Problem Statement
Consider a producer supplies batches of articles having mean value 
of a variable (length, weight, strength, etc.) and standard deviation 
of the variable. The consumer has agreed that a batch will be
acceptable if
0  T     0  T
where 0 denotes the critical (nominal) mean value of the variable
and T indicates the tolerance for the mean value of the variable.
Otherwise, the batch will be rejected.
Here, the producer’s risk  is the probability of rejecting a perfect
batch, the one for which = 0 .
The consumer’s risk  is the probability of accepting an imperfect
batch, the one for which = 0 T.
Sampling Scheme
Let us assume that the probability distribution of mean x of samples,
each of size n, taken from a batch, is (or tends to) normal with mean
, where  =0 and standard deviation  n . Then, the batch will be
accepted if
0  t  x  0  t
where t denotes the tolerance for sample mean x . Otherwise, the
batch will be rejected.
Here, the producer’s risk  is the probability of rejecting a perfect
batch and the consumer’s risk  is the probability of accepting an
imperfect batch.
The Producer’s Risk Condition
2
2
0  T
0  t  0 0  t
u 2
0  t   0



n
0  T
t n


where u 2 is the standard normal variable
corresponding to a tail area  2.
x
The Consumer’s Risk Condition
 2
 2
0  T 0  t
u 2
0
 0  t 0  T
 0  t     0  T  T  t 




n
n

where u 2
is the standard normal variable
corresponding to a tail area  2.
x
Finding n and t
u 2
T t


n

 u
u 2
t n

;

or, t  
u 2
n
2
2
2
2

n    u 2  u 2  T 2
2
or, n    u 2  u 2  T 2
T n t n T n



 u 2 ;



u 2
2  u 2  u 2  T 2
2
t  Tu 2
 Tu 2
u
 2
u
 u 2 
 2
 u 2 
Illustration
A producer (spinner) supplies yarn of nominal linear density equal to
be 45 tex. The customer (knitter) accepts yarn if its mean linear
density lies within a range of 451.5 tex. As the knitter cannot test all
the yarns supplied by the spinner, the knitter would like to devise an
acceptance sampling scheme with 10% producer’s risk and 5%
consumer’s risk . Assume the standard deviation of count within a
delivery is 1.2 tex.
Here 0tex  45, Ttex  1.5,   0.10,   0.05, tex  1.2
Assume the mean linear density of yarn samples, each of size n,
follows (or tends to follow) normal distribution with mean 45 tex and
standard deviation 1.2 tex. Then, the standard normal variable takes
the following values
u 2  u0.05  1.6449
u 2  u0.025  1.9600
Illustration…
Then, n  
2
 tex 
and
u
 2 
ttex   Ttex u 2
 u 2
u
 2 

2
2
 tex 
T
 1.2  1.6449  1.9600 
2
2
1.5
2
 8.3  9

 u 2  1.51.6449  1.6449  1.9600   0.68
Thus, the sampling scheme is as follows: Take a yarn sample of size
nine and accept the delivery if the sample mean lies in the range of
450.68 tex, that is in-between 44.32 tex and 45.68 tex, otherwise,
reject the delivery.
Frequently Asked Questions & Answers
Frequently Asked Questions & Attributes
Q1: How the practical OC curve can be made closer to the ideal OC curve?
A1: By increasing the sample size, the practical OC curve can be made closer to
the ideal OC curve.
Q2: How the discriminatory power of the acceptance sampling plan can be
increased?
A2: The discriminatory power of the acceptance sampling plan can be increased
by increasing the sample size. Also, plans with smaller value of c provide
discrimination at lower levels of lot fraction defective than do plans with larger
values of c.
Q3: What are the specific points of OC curve that a quality engineer looks for?
A3: A quality engineer always looks for the AQL and the LTPD in an OC curve.
References
1.
Gupta, S. C. and Kapoor, V. K., Fundamentals of Applied Statistics, Sultan
Chand & Sons, New Delhi, 2007.
2.
Leaf, G. A. V., Practical Statistics for the Textile Industry: Part II, The Textile
Institute, UK, 1984.
Sources of Further Reading
1.
Leaf, G. A. V., Practical Statistics for the Textile Industry: Part I, The Textile
Institute, UK, 1984.
2.
Leaf, G. A. V., Practical Statistics for the Textile Industry: Part II, The Textile
Institute, UK, 1984.
3.
Gupta, S. C. and Kapoor, V. K., Fundamentals of Applied Statistics, Sultan
Chand & Sons, New Delhi, 2007.
4.
Gupta, S. C. and Kapoor, V. K., Fundamentals of Mathematical Statistics,
Sultan Chand & Sons, New Delhi, 2002.
5.
Montgomery, D. C., Introduction to Statistical Quality Control, John Wiley &
Sons, Inc., Singapore, 2001.
6.
Grant, E. L. and Leavenworth, R. S., Statistical Quality Control, Tata McGraw
Hill Education Private Limited, New Delhi, 2000.
7.
Montgomery, D. C. and Runger, G. C., Applied Statistics and Probability for
Engineers, John Wiley & Sons, Inc., New Delhi, 2003.